Secondary 1 Science Semestral Assessment 2 (End of Year) Paper 4

TuitionGoWhere Practice Paper - Science Secondary 1

SA2 Version 4 - Answer Key and Marking Scheme

Total Marks: 60

Section A: Multiple Choice Questions [10 marks]

Question 1 [1 mark]

Answer: C

Working:

• Weight of book = mg = 2 kg × 10 N/kg = 20 N
• Work done against gravity = Force × distance = 20 N × 1.5 m = 30 J

Key concept: Work done against gravity = mgh when lifting at constant velocity.

Question 2 [1 mark]

Answer: A

Explanation: A compressed spring stores elastic potential energy. When released, this energy is converted into kinetic energy of the toy car (assuming negligible friction and no height change).

Common mistake: Confusing with gravitational potential energy (C) or chemical energy (B).

Question 3 [1 mark]

Answer: B

Working:

• Net force = Applied force - Friction = 20 N - 8 N = 12 N
• Net work done = Net force × distance = 12 N × 4 m = 48 J

Alternative method:

• Work by applied force = 20 N × 4 m = 80 J
• Work against friction = 8 N × 4 m = 32 J
• Net work = 80 J - 32 J = 48 J

Question 4 [1 mark]

Answer: B

Working:

• Initial kinetic energy = ½mv² = ½ × 0.5 kg × (10 m/s)² = 25 J
• At maximum height, all KE is converted to GPE (conservation of energy)
• Maximum GPE = 25 J

Key concept: For vertical projection ignoring air resistance, max GPE = initial KE.

Question 5 [1 mark]

Answer: B

Explanation: Power is defined as the rate of doing work, or work done per unit time. Unit: Watt (W) = Joule/second (J/s).

Common mistakes:

• A: Confuses power with work
• C: Joule is unit of work/energy, not power
• D: Power decreases when time increases for same work (P = W/t)

Question 6 [1 mark]

Answer: C

Working:

• Work done = Force × distance = 2000 N × 15 m = 30,000 J
• Power = Work / time = 30,000 J / 30 s = 1000 W

Question 7 [1 mark]

Answer: A

Explanation: On a frictionless inclined plane, GPE decreases linearly with distance (GPE = mgh, h decreases linearly with distance). By conservation of energy, KE increases by the same amount, so KE increases linearly with distance.

Question 8 [1 mark]

Answer: C

Working:

• Work done by student = Applied force × distance
• 120 J = F × 6 m
• F = 20 N

Check: Net force = 20 N - 10 N = 10 N, Net work = 10 N × 6 m = 60 J (goes to KE)

Question 9 [1 mark]

Answer: A

Explanation: Work done = Force × distance moved in direction of force. When holding a barbell stationary, there is no displacement, so work done = 0 J. The weightlifter exerts an upward force but the barbell does not move.

Key concept: No displacement → no work done, regardless of force magnitude.

Question 10 [1 mark]

Answer: B

Working:

• Work done = Change in kinetic energy (Work-Energy Theorem)
• Initial KE = ½ × 2 kg × (5 m/s)² = 25 J
• Final KE = ½ × 2 kg × (9 m/s)² = 81 J
• Work done = 81 J - 25 J = 56 J

Section B: Structured Questions [30 marks]

Question 11 [4 marks]

(a) Net force = 25 N - 7 N = 18 N [1]

• Direction: Same as applied force (to the right)

(b) Acceleration = Net force / mass = 18 N / 3 kg = 6 m/s² [1]

(c) Work done by applied force = Force × distance = 25 N × 6 m = 150 J [1]

(d) Final kinetic energy = Net work done = Net force × distance = 18 N × 6 m = 108 J [1]

• Alternative: Using v² = u² + 2as = 0 + 2(6)(6) = 72, KE = ½mv² = ½(3)(72) = 108 J

Marking notes:

• (a) Must subtract friction correctly
• (b) Use F=ma with net force
• (c) Use applied force only, not net force
• (d) Work-energy theorem: net work = ΔKE

Question 12 [5 marks]

(a) Principle of conservation of energy: Energy cannot be created or destroyed, only converted from one form to another. The total energy in a closed system remains constant. [1]

(b) GPE at A = mgh = 500 kg × 10 N/kg × 40 m = 200,000 J [1]

(c) At point B (ground level), GPE = 0. By conservation of energy:

• KE at B = GPE at A = 200,000 J
• ½mv² = 200,000 J
• ½ × 500 × v² = 200,000
• v² = 800
• v = √800 = 28.3 m/s (or 20√2 m/s) [2]

Mark breakdown: 1 mark for correct equation/setup, 1 mark for correct answer with unit.

(d) At point C:

• GPE at C = mgh = 500 × 10 × 25 = 125,000 J
• KE at C = Total energy - GPE at C = 200,000 - 125,000 = 75,000 J
• ½ × 500 × v² = 75,000
• v² = 300
• v = √300 = 17.3 m/s (or 10√3 m/s) [1]

Question 13 [4 marks]

(a) Graph plotting [1]

• Axes labelled correctly with units (Force/N vertical, Extension/cm horizontal)
• Appropriate scale using >50% of grid
• All 6 points plotted accurately
• Best-fit straight line through origin

(b) The extension of the spring is directly proportional to the force applied (Hooke's Law). [1]

• Or: Force is directly proportional to extension.

(c) Spring constant k = Force / Extension [2]

• From graph: gradient = (10 N - 0 N) / (15 cm - 0 cm) = 10/15 N/cm = 2/3 N/cm
• Convert to N/m: k = (2/3) × 100 = 66.7 N/m
• Or using any point: k = 10 N / 0.15 m = 66.7 N/m

Mark breakdown: 1 mark for correct method (gradient or F/x), 1 mark for correct answer with unit (N/m).

Common mistake: Forgetting to convert cm to m, giving answer in N/cm.

Question 14 [5 marks]

(a) Work done against gravity = mgh = 12 kg × 10 N/kg × 8 m = 960 J [1]

(b) Useful power output = Work / time = 960 J / 10 s = 96 W [1]

(c) Electrical energy input = VIt = 240 V × 1.5 A × 10 s = 3600 J [1]

(d) Efficiency = (Useful energy output / Energy input) × 100% [2]

• = (960 J / 3600 J) × 100% = 26.7%
• Or using power: Efficiency = (96 W / 360 W) × 100% = 26.7%

Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with % sign.

Question 15 [4 marks]

(a) GPE lost = mgh = 0.2 kg × 10 N/kg × 0.15 m = 0.3 J [1]

(b) Kinetic energy at lowest point = GPE lost = 0.3 J [1]

• By conservation of energy (ignoring air resistance)

(c) KE = ½mv² [1]

• 0.3 = ½ × 0.2 × v²
• v² = 3
• v = √3 = 1.73 m/s

(d) In a real situation, air resistance and friction at the pivot do negative work on the bob, converting some mechanical energy to thermal energy/sound. Thus, the bob has less kinetic energy at the bottom and cannot rise to the same height. [1]

Key concept: Non-conservative forces dissipate mechanical energy.

Question 16 [4 marks]

(a) Vertical height = 25 steps × 0.18 m/step = 4.5 m [1]

(b) Work done against gravity = mgh = 60 kg × 10 N/kg × 4.5 m = 2700 J [1]

(c) Power = Work / time = 2700 J / 6 s = 450 W [1]

(d) Efficiency = Useful energy output / Chemical energy input [1]

• 0.25 = 2700 J / Chemical energy input
• Chemical energy input = 2700 J / 0.25 = 10,800 J

Key concept: Human body efficiency ~25%, most chemical energy becomes heat.

Question 17 [4 marks]

(a) Gain in GPE = mgh = 5 kg × 10 N/kg × 1.5 m = 75 J [1]

(b) Work done by applied force = Force × distance along plane = 40 N × 4 m = 160 J [1]

(c) Work done against friction = Work by applied force - Gain in GPE [1]

• = 160 J - 75 J = 85 J
• (Since constant velocity, ΔKE = 0, so net work = 0)

(d) Work against friction = Friction × distance [1]

• 85 J = Friction × 4 m
• Friction = 21.25 N

Alternative for (d): Since constant velocity, net force parallel to plane = 0

• Applied force = Friction + mg sinθ
• mg sinθ = 5×10×(1.5/4) = 18.75 N
• Friction = 40 - 18.75 = 21.25 N

Section C: Longer Structured Questions [20 marks]

Question 18 [7 marks]

(a) Elastic potential energy = ½kx² [2]

• = ½ × 200 N/m × (0.1 m)²
• = ½ × 200 × 0.01
• = 1 J

Mark breakdown: 1 mark for correct formula, 1 mark for correct substitution and answer with unit.

(b) Kinetic energy just after launch = Elastic potential energy = 1 J [1]

• Assuming no energy losses in spring/launch mechanism

(c) At maximum height, KE = 0, all initial energy converted to GPE [2]

• GPE = mgh = 1 J
• 0.5 kg × 10 N/kg × h = 1 J
• h = 1 / 5 = 0.2 m

Mark breakdown: 1 mark for energy conservation equation, 1 mark for correct answer with unit.

(d) Distance up ramp = h / sinθ = 0.2 m / sin30° = 0.2 / 0.5 = 0.4 m [1]

(e) If friction is present, some mechanical energy is converted to thermal energy/sound. The car would have less kinetic energy at the bottom of the ramp, so it would reach a lower maximum height. [1]

Question 19 [7 marks]

(a) GPE lost per second = mass flow rate × g × h [2]

• = 500 kg/s × 10 N/kg × 80 m
• = 400,000 J/s = 400,000 W

Mark breakdown: 1 mark for correct formula (mgΔh per second), 1 mark for correct answer with unit (W or J/s).

(b) Electrical power output = Efficiency × Input power [2]

• = 0.85 × 400,000 W
• = 340,000 W = 340 kW

Mark breakdown: 1 mark for correct use of efficiency, 1 mark for correct answer with unit.

(c) Two other forms of energy: [2]

1. Thermal energy (heat) due to friction in turbines, generators, and water turbulence
2. Sound energy from moving water and machinery

• (Also acceptable: Kinetic energy of water leaving the turbine)

Mark breakdown: 1 mark each for two valid forms.

(d) Increase the height of the water fall (increase head height). [1]

• Or: Increase the efficiency of the turbine/generator system.
• Or: Use water with higher density (not practical).

Key concept: Power output = ηρgQh, so increasing h increases power.

Question 20 [6 marks]

(a) At constant velocity, acceleration = 0. By Newton's First Law, net force = 0. [2]

• Horizontal forces: Applied force (spring balance reading) to the right, Friction to the left
• Since net force = 0, Applied force = Friction
• Therefore, spring balance reading = magnitude of frictional force

Mark breakdown: 1 mark for net force = 0 at constant velocity, 1 mark for equating spring balance reading to friction.

(b) Work done against friction = Friction × distance = 6 N × 3 m = 18 J [1]

(c) With 1 kg mass added: [2]

• Total mass = 2 kg + 1 kg = 3 kg
• Normal reaction = Weight = 3 kg × 10 N/kg = 30 N
• Frictional force = Spring balance reading = 9 N
• Coefficient of friction μ = Friction / Normal reaction = 9 N / 30 N = 0.3

Mark breakdown: 1 mark for correct normal reaction, 1 mark for correct μ calculation.

(d) The graph of frictional force against normal reaction is a straight line passing through the origin. [1]

• The gradient represents the coefficient of friction (μ).

Key concept: For kinetic friction, F_f = μR, so F_f vs R is linear through origin with gradient μ.

End of Answer Key